Magnetism in low dimensions from first principles. Atomic magnetism. Gustav Bihlmayer. Gustav Bihlmayer

Transcription

1 IFF 10 p. 1 Magnetism in low dimensions from first principles Atomic magnetism Gustav Bihlmayer Institut für Festkörperforschung, Quantum Theory of Materials Gustav Bihlmayer Institut für Festkörperforschung Institute for Advanced Simulation Forschungszentrum Jülich

2 IFF 10 p. 2 Atomic-like states in oxides Magnetite: Fe3O4 Fe 3+ Fe 3+ Fe O Oxygen (fcc) tetrahedral Fe octahedral Fe magnetic moment: A2 (here) influence of crystal field: A3 (next talk) electronic & magnetic interaction: A4 and A6 (tomorrow) reduced hopping between Fe sites: - atomic-like behaviour (A2) - correlation effects (A5, A8...)

3 IFF 10 p. 2 Atomic-like states in oxides 2+ Fe in octahedral field 6 d-electrons 3+ Fe in tetrahedral field t 2g 5 d-electrons e g electronic levels filled according to Hund s rules

4 IFF 10 p. 3 Electronic states in an atom Intraatomic exchange interaction Atoms in a magnetic field Spin-Orbit coupling Hund s rules in a solid

5 IFF 10 p. 4 Atomic Schrödinger equation Non-relativistic, N particles, nuclear number Z: i i Z r i j i 1 r ij Ψ(r 1,r 2,...,r N ) = EΨ(r 1,r 2,...,r N ) Rotationally invariant Hamiltonian: angular momentum (L) is a good quantum number. orbital (angular) momentum operator: L = i h(r ) Eigenvalues of L 2 : h 2 L(L + 1) Eigenvalues of L z : hm L

6 IFF 10 p. 5 Single electron quantum numbers single electron with spin σ : ( 1 2 Z r ) ψ(r) = Eψ(r) Transformation r r, ϑ, φ Separation: ψ(r) = R n,l (r)θ l,ml (ϑ)φ ml (φ)σ ms (σ) Principal quantum number n defines energy eigenvalue Azimuthal quantum number l orbital angular momentum Magnetic quantum number m l orientation of l w.r.t. an axis Spin quantum number m s spin (m s = ± 1 2 ) w.r.t. an axis

7 IFF 10 p. 6 Many (two) electron states Single electron states m (1) l, m (1) s and m (2) l, m (2) s E.g. 2p states: l = 1 therefore m l = 1, 0, 1 and m s =,

8 IFF 10 p. 6 Many (two) electron states Single electron states m (1) l, m (1) s and m (2) l, m (2) s E.g. 2p states: l = 1 therefore m l = 1, 0, 1 and m s =, Possible 2-electron states: 15 determinants m (1) l m (1) s, m (2) l m (2) s = 15 (Pauli principle, indistinguishable particles)

9 IFF 10 p. 6 Many (two) electron states Single electron states m (1) l, m (1) s and m (2) l, m (2) s E.g. 2p states: l = 1 therefore m l = 1, 0, 1 and m s =, Possible 2-electron states: 15 determinants m (1) l m (1) s, m (2) l m (2) s = 15 (Pauli principle, indistinguishable particles) Group according to M L = i m(i) l E.g. 1, 1 has M L = 2 and M S = 0 and M S = i m(i) s

10 IFF 10 p. 6 Many (two) electron states Group according to M L = i m(i) l E.g. 1, 1 has M L = 2 and M S = 0 and M S = i m(i) s 1,1 1 D M L 1,0 1,0 1,0 1,0 1, 1 1, 1 0,0 1, 1 1, 1 1 S 0, 1 0, 1 0, 1 0, 1 1, 1 3 P M S

11 IFF 10 p. 7 Electronic multiplets Multiplet term symbol: M L where M = 2M S + 1 and L = S, P, D, F,... for 2p states: 1 D, 3 P, 1 S. Other example with d-states: 3d 2 1 G 1 G 3 F 1 D 3 P M L 1 S density matrix (m,m ) 3F 2 m m 1 1 G D F M S 3F 4

12 IFF 10 p. 8 Hund s rules Which multiplet is the ground state? select largest spin multiplicity M select largest orbital multiplicity L compatible with M less than half-filled shells: M S antiparallel to M L more than half-filled shells: M S parallel to M L

13 IFF 10 p. 9 Electronic states in an atom Intraatomic exchange interaction Atoms in a magnetic field Spin-Orbit coupling Hund s rules in a solid

14 IFF 10 p. 10 Example: 2 particles Schrödinger equation (H 1 + H 2 + H 12 )Ψ(r 1,r 2 ) = EΨ(r 1,r 2 ) where H 12 = 1 r 12 and H i = i Z r i with H i ψ ν (r i ) = ǫ ν ψ ν (r i ) Ansatz for Ψ (spatial part): Ψ(r 1,r 2 ) = aψ 1 (r 1 )ψ 2 (r 2 ) + bψ 1 (r 2 )ψ 2 (r 1 ) = a b

15 IFF 10 p. 10 Example: 2 particles Schrödinger equation (H 1 + H 2 + H 12 )Ψ(r 1,r 2 ) = EΨ(r 1,r 2 ) where H 12 = 1 r 12 Ansatz for Ψ (spatial part): Ψ(r 1,r 2 ) = aψ 1 (r 1 )ψ 2 (r 2 ) + bψ 1 (r 2 )ψ 2 (r 1 ) = a b leads to H 12 (a b ) = (E ǫ 1 ǫ 2 )(a b ) = ǫ 12 (a b ) or a( H ǫ 12 ) + b H = 0 a H b( H ǫ 12 ) = 0

16 IFF 10 p. 11 Coulomb & exchange integrals Coulomb integral: J = H = ψ 1 (r 1 ) 2 ψ 2 (r 2 ) 2 r 12 dr 1 dr 2 Exchange integral: K = H = ψ1 (r 1)ψ2 (r 2)ψ 1 (r 2 )ψ 2 (r 1 ) r 12 dr 1 dr 2 rewrite in matrix form: ( (J ǫ 12 ) K K (J ǫ 12 ) ) ( a b ) = 0

17 IFF 10 p. 11 Coulomb & exchange integrals ( (J ǫ 12 ) K K (J ǫ 12 ) ) ( a b ) = 0 has solutions ǫ 12 = J K : a = b = 1 2 and ǫ 12 = J+K : a = b = 1 2 Since K > 0, for the lowest energy the spatial part of Ψ is antisymmetric, therefore the spin part has to be symmetric: 1 2 or 1 2 or This describes a triplet state.

18 IFF 10 p. 12 Example: the He atom Both electrons in 1s level: Spin part has to be antisymmetric: ) 1 2 E 1 = 2ǫ 1s + J + K Ψ 1 = 1s 1 1s 2 (

19 IFF 10 p. 12 Example: the He atom Both electrons in 1s level: Spin part has to be antisymmetric: ) 1 2 E 1 = 2ǫ 1s + J + K Ψ 1 = 1s 1 1s 2 ( Electrons in 1s and 2s level: Spin part can be symmetric (triplet): E 2 = ǫ 1s +ǫ 2s +J K Ψ 2 = 1 1s 1 2s 2 1s 2 2s

20 IFF 10 p. 12 Example: the He atom Both electrons in 1s level: Spin part has to be antisymmetric: ) 1 2 E 1 = 2ǫ 1s + J + K Ψ 1 = 1s 1 1s 2 ( Electrons in 1s and 2s level: Spin part can be symmetric (triplet): E 2 = ǫ 1s +ǫ 2s +J K Ψ 2 = 1 1s 1 2s 2 1s 2 2s E 3 = ǫ 1s + ǫ 2s + J + K singlet is higher in energy

21 IFF 10 p. 13 Two p-electrons carbon atom: configuration 1s 2 2s 2 2p 2 ; possible multiplets: 1 D, 3 P, 1 S 1 D term: M L = 2, M S = 0, e.g. E( 1 D) = E( 1, 1 ) 3 P term: M L = 1, M S = 1, e.g. E( 3 P) = E( 1, 0 ) 1 S term: M L = 0, M S = 0 (contains also 1 D and 3 P ). 1,1 1 D M L 1,0 1,0 1,0 1,0 1, 1 1, 1 0,0 1, 1 1, 1 1 S 0, 1 0, 1 0, 1 0, 1 1, 1 3 P M S

22 IFF 10 p. 14 Coulomb & exchange integrals (2) Problem: compare different Coulomb & exchange integrals Simplification: factorize (atomic-like) wavefunctions: Ψ α (r 1 ) = 1 r 1 R(r 1 ; n α l α )Θ(ϑ 1 ; l α m α l )Φ(φ 1 ; m α l )Σ(σ 1 ; m α s ) rewrite J and K in angular (a, b) and radial (F, G) part: α 1 β 2 1 α 1 β 2 r 12 α 1 β 2 1 β 1 α 2 r 12 = = a (k) (l α m α l, l β m β l )F (k) (n α l α, n β l β ) k=0 b (k) (l α m α l, l β m β l )G(k) (n α l α, n β l β ) k=0

23 Ordering of the multiplets With the help of these Slater parameters, F (k) J.C.Slater, Phys.Rev. 34, 1293 (1929) we can write the energies of the terms: E( 1 S) = 2E 2p + F F 2 E( 1 D) = 2E 2p + F F 2 E( 3 P) = 2E 2p + F F 2 8. Mar Since F 0 = F 0 (21, 21) > 0 and F 2 = F 2 (21, 21) > 0, the triplet is again the ground state. IFF 10 p. 15

24 IFF 10 p. 16 Electronic states in an atom Intraatomic exchange interaction Atoms in a magnetic field Spin-Orbit coupling Hund s rules in a solid

25 IFF 10 p. 17 Schrödinger equation + magnetic field total momentum of charged particle in external field: p q c A (minimal coupling) interaction of a magnetic moment µ with a magnetic field: V Z = µ B where µ = µ B g s S For an electron of unit charge and g s = 2, we can introduce Lorentz-force and Zeeman term in the Schrödinger equation: i ( 1 2 ( p(r i ) 1 ) ) 2 c A(r i) + V (r i ) + 1 2c σ i B(r i ) j i 1 r ij Ψ = EΨ

26 IFF 10 p. 18 Magnetic field and vector potential Vector potential A and magnetic field: B = A Coulomb gauge: A = 0 Magnetic field in z-direction: A = 1 2 r 0 0 B z = 1 2 yb z xb z 0 therefore ( p 1 c A ) 2 = p 2 + 2i c A + 1 c 2A2 = p c B L + 1 4c 2B2 z(x 2 + y 2 )

27 IFF 10 p. 19 Perturbation theory Extract magnetic part of Hamiltonian H mag = 1 2c B (L i + σ i ) + 1 i 8c 2B2 z (x 2 i + yi 2 ) i and consider these terms as a perturbation: E = E (0) H mag 0 + n 0 first order in B: paramagnetic term 0 H mag n 2 E (0) 0 E n (0) second order in B: diamagnetic and van-vleck paramagnetic term

28 IFF 10 p. 20 Paramagnetism With µ B = 1 2c and µ 0 tot = µ B 0 i (L i + 2S i ) 0 first order term: µ B B 0 L i + 2S i 0 = B µ 0 tot i exists only, if there is a moment in the ground state! magnet liquid oxygen for 1µ B and 1 Tesla 10 5 ev paramagnetism of oxygen: from demo.physics.uiuc.edu

29 IFF 10 p. 21 Diamagnetism + 1 8c 2B2 z 1 8c 2B2 z 0 i 0 i (x 2 i + yi 2 ) 0 ri 2 sinϑ 2 i 0 = induced moment antiparallel to field exists also without permanent moment! magnitude ev levitation of pyrolytic carbon in an external field (from en.wikipedia.org)

30 Paramagnet in a B-field MS = +3/2 Without orbital moment: 2µ B B 0 S i 0 = i 2µ B B z 0 S z,i 0 i Level splitting according to M S! with orbital moment: total angular momentum J = L+S S = 3/2 without field E 4 S MS = +1/2 MS = -1/2 MS = -3/2 with field How do L and S couple? 8. Mar B IFF 10 p. 22

31 IFF 10 p. 23 Electronic states in an atom Intraatomic exchange interaction Atoms in a magnetic field Spin-Orbit coupling Hund s rules in a solid

32 IFF 10 p. 24 Lorentz Transformation Electric and magnetic fields in a moving reference-frame (primed): E = γ (E v B) B = γ 1 with γ = 1 v2 /c 2 ( B v E ) c 2 ( ) γ 1 (E v)v v 2 ( ) γ 1 (B v)v For v << c and B = 0 we get B 1 c 2 (v E) v 2

33 IFF 10 p. 25 The spin-orbit coupling (SOC) term Realtivistic effects: Dirac equation! Up to order (v/c) 2 : [ E V (r) c 2 (E V (r))2 i c A(r) 1 2c 2A2 (r) + i (2c) 2E(r) p 1 (2c) 2 σ (p E(r)) + 1 2c σ B(r) ] ψ = 0 potential near the nucleus B p σ electron in an atom E

34 IFF 10 p. 26 Atomic SOC σ (E(r) p) = σ ( V (r) p) = 1 r dv (r) σ (r p) = 1 dr r dv (r) (σ L) dr coupling between spin and orbital momentum spin and orbital moment are antiparallel Coulomb-like potential for small r like V (r) = Z r ξ = 1 r dv (r) dr Z coupling strength scales with ψ ξ ψ, i.e. approx. Z 2 large for heavy atoms, splittings up to 1.5 ev (Bi)

35 IFF 10 p. 27 Hydrogenic atom with SOC E = E NR + E soc = 1 2 Z 2 n 2 α2 Z 4 2n 3 ( 1 j ) 4n n l s j E NR [ Z2 2 ] E soc[ α2 Z 4 2 ] 1 0 1/2 1/2 1 1/4 0 n = 3 n = 2 3D 5/2 3P 3/2,3D 3/2 3S 1/2,3P 1/ /2 1/2 1/4 5/64 2P 3/2 2S 1/2,2P 1/ /2 1/2 1/4 5/ /2 1/2 1/4 1/ /2 1/2 1/9 9/ /2 1/2 1/9 9/ /2 3/2 1/9 3/324 n = /2 3/2 1/9 3/ /2 5/2 1/9 1/324 1S 1/2

36 IFF 10 p. 28 More than 1 electron & SOC Small SOC: J = S + L with S = i s i and L = i l i Large SOC: J = i j i with j i = s i + l i µ tot = µ orb + µ spin = µ B (L + 2S) total moment is not collinear to J

37 IFF 10 p. 28 More than 1 electron & SOC Small SOC: J = S + L with S = i s i and L = i l i Large SOC: J = i j i with j i = s i + l i µ tot = µ orb + µ spin = µ B (L + 2S) total moment is not collinear to J MJ = +3/2 Small B-field: MJ = +1/2 µ tot = g J µ B J(J + 1) J = 3/2 MJ = -1/2 (µ tot ) z = g J µ B M J with g J = 1 + J(J+1)+S(S+1) L(L+1) 2J(J+1) without field MJ = -3/2 with field E=g µ B J B

38 IFF 10 p. 29 Strong magnetic field Single electron, weak field: strong field: E B = g J µ B m J B z E B = µ B (m l + 2m s )B z Example: 3 P 3/2 : m l m s m j m l + 2m s 1 1/2 3/ /2 1/ /2 1/ /2 3/2 2 (Paschen-Beck effect)

39 IFF 10 p. 30 Electronic states in an atom Intraatomic exchange interaction Atoms in a magnetic field Spin-Orbit coupling Hund s rules in a solid

40 IFF 10 p. 31 Hund s rules (1) select largest spin multiplicity M (2) select largest orbital multiplicity L compatible with M (3a) less than half-filled shells: M S antiparallel to M L, i.e. J = L S (3b) more than half-filled shells: M S parallel to M L, i.e. J = L + S (1) and (2) are Coulomb (and exchange) effects (3a) and (3b) follow from spin-orbit coupling S=1L=-5 S=5/2 L=0 S=3/2 L=5

41 IFF 10 p. 31 Hund s rules (1) select largest spin multiplicity M (2) select largest orbital multiplicity L compatible with M (3a) less than half-filled shells: M S antiparallel to M L, i.e. J = L S (3b) more than half-filled shells: M S parallel to M L, i.e. J = L + S Ions in a solid : rule (1) survives orbital moments get quenched by the crystal field if sizable orbital moments arise, rule (3) holds

42 IFF 10 p. 32 Example: free rare earth (RE 3+ ) ions 8 6 S L J La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu

43 IFF 10 p. 33 RE 3+ ions in a solid: garnets RE 3 Fe 5 O 12, better: 3 RE 2 O 3 5 Fe 2 O 3 Fe 3+ atoms: 10µ B RE 3+ ions: 6(L + 2S)µ B couple antiferromagnetically

44 IFF 10 p. 33 RE 3+ ions in a solid: garnets RE 3 Fe 5 O 12, better: 3 RE 2 O 3 5 Fe 2 O 3 Fe 3+ atoms: 10µ B RE 3+ ions: 6(L + 2S)µ B couple antiferromagnetically mag. moment (µ B ) exp. moment 12S (L+2S)-10 6(L/3+2S)-10 Y Gd Tb Dy Ho Er Tm Yb Lu L is quenched in the solid, but S survives!

45 IFF 10 p. 34 Once more: magnetite (Fe 2+ O 2 )(Fe 3+ 2 O2 3 ) small µ orb anisotropy = J/m 3 d m = L "e g " e g t 2g "t " 2g m = 0 L cubic trigonal spherical atomic spin-orbit splitting Fe 2+ Fe 2+ Fe 2+

46 IFF 10 p. 34 Once more: magnetite (Fe 2+ O 2 )(Fe 3+ 2 O2 3 ) small µ orb anisotropy = J/m 3 (Co 2+ O 2 )(Fe 3+ 2 O2 3 ) large µ orb anisotropy 10 6 J/m 3 d m = L "e g " e g t 2g "t " 2g m = 0 L cubic trigonal spherical atomic spin-orbit splitting Co 2+ Co 2+ Co 2+

47 IFF 10 p. 35 Summary transition metal atoms in oxides carry an atomic-like spin-moment S but have a quenched orbital moment L that couples due to SOC to form a total angular moment J these moments behave according to Hund s rules first maximizing the spin multiplicity then maximizing a compatible orbital moment total moment determined by SOC depending of filling: J = L S or J = L + S talk available at: gbihl/publ/g_bihlmayer_pub.html#b016